How do you find the radius of convergence Sigma (n!x^n)/sqrt(n^n) from n=[1,oo)?

1 Answer
mason m
Mar 2, 2017

The interval of convergence is x=0 and the radius of convergence is R=0.

Explanation:

sum_(n=1)^oo(n!(x^n))/sqrt(n^n)=sum_(n=1)^oo(n!(x^n))/n^(n/2)

The series suma_n is convergent when L<1 when L=lim_(nrarroo)abs(a_(n+1)/a_n). Here, this becomes:

L=lim_(nrarroo)abs(((n+1)!(x^(n+1)))/(n+1)^(1/2(n+1))*n^(n/2)/(n!(x^n)))

Simplifying:

L=lim_(nrarroo)abs(((n+1)(n!))/(n!)*x^(n+1)/x^n*n^(n/2)/(n+1)^(n/2+1/2))

L=lim_(nrarroo)abs((n+1)/(n+1)^(n/2+1/2)*x*n^(n/2))

The x can be moved to outside the limit, since the limit only depends on n.

Also note that (n+1)/(n+1)^(n/2+1/2)=(n+1)^(1-(n/2+1/2))=(n+1)^(1/2-n/2).

L=absxlim_(nrarroo)abs((n+1)^(1/2-n/2)(n^(n/2)))

L=absx(lim_(nrarroo)abs((n+1)(n/(n+1))^n))^(1/2)

Note that lim_(nrarroo)((n+1)/n)=e, so lim_(nrarroo)(n/(n+1))=1/e. Thus, the overall limit approaches oo since the remaining portion of n+1 is unbounded.

Since the limit approaches infinity, the only time when L<1 will be when x=0, as this makes L=0.

Thus the interval of convergence is x=0 and the radius of convergence is R=0.

Related questions
See all questions in Determining the Radius and Interval of Convergence for a Power Series
Impact of this question
2697 views around the world
You can reuse this answer
Creative Commons License